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प्रश्न
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (APB) + ar (PCD) = 1/2ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
[Hint: Through. P, draw a line parallel to AB]
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उत्तर
(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.
In parallelogram ABCD,
AB || EF (By construction) ... (1)
ABCD is a parallelogram.
∴ AD || BC (Opposite sides of a parallelogram)
⇒ AE || BF ... (2)
From equations (1) and (2), we obtain
AB || EF and AE || BF
Therefore, quadrilateral ABFE is a parallelogram.
It can be observed that ΔAPB and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF.
∴ Area (ΔAPB) = 1/2Area (ABFE) ... (3)
Similarly, for ΔPCD and parallelogram EFCD,
Area (ΔPCD) = 1/2Area (EFCD) ... (4)
Adding equations (3) and (4), we obtain
Area (ΔAPB) + Area (ΔPCD) = 1/2[Area (ABFE) + Area (EFCD)]
Area (ΔAPB) + Area (ΔPCD) = 1/2Area (ABCD) ........(5)
(ii)
Let us draw a line segment MN, passing through point P and parallel to line segment AD.
In parallelogram ABCD,
MN || AD (By construction) ... (6)
ABCD is a parallelogram.
∴ AB || DC (Opposite sides of a parallelogram)
⇒ AM || DN ... (7)
From equations (6) and (7), we obtain
MN || AD and AM || DN
Therefore, quadrilateral AMND is a parallelogram.
It can be observed that ΔAPD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.
∴ Area (ΔAPD) = 1/2Area (AMND) ... (8)
Similarly, for ΔPCB and parallelogram MNCB,
Area (ΔPCB) = 1/2Area (MNCB) ... (9)
Adding equations (8) and (9), we obtain
Area (ΔAPD) + Area (ΔPCB) = 1/2[Area (AMND) + Area (MNCB)]
Area (ΔAPD) + Area (ΔPCB) = 1/2Area (ABCD) ...........(10)
On comparing equations (5) and (10), we obtain
Area (ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area (ΔPCD)
संबंधित प्रश्न
In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2ar (PQRS)
In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that
ar (BPC) = ar (DPQ).
[Hint: Join AC.]
In the given below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar (ΔADE)
= ar (ΔBCF)
ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC
such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
(1) ar ( ADEG) = ar (GBCD)
(2) ar (ΔEGB) = `1/6` ar (ABCD)
(3) ar (ΔEFC) = `1/2` ar (ΔEBF)
(4) ar (ΔEBG) = ar (ΔEFC)
(5)ΔFind what portion of the area of parallelogram is the area of EFG.
In which of the following figures, you find two polygons on the same base and between the same parallels?
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm2.
ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF (Figure), prove that ar (AER) = ar (AFR)
In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Figure). Prove that ar (ABCD) = ar (APQD)
The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.
